How do I catch “Index out of range” in Swift?

As suggested in comments and other answers it is better to avoid this kind of situations. However, in some cases you might want to check if an item exists in an array and if it does safely return it. For this you can use the below Array extension for safely returning an array item.

Swift 5

extension Collection where Indices.Iterator.Element == Index {
    subscript (safe index: Index) -> Iterator.Element? {
        return indices.contains(index) ? self[index] : nil
    }
}

Swift 4

extension Collection where Indices.Iterator.Element == Index {
    subscript (safe index: Index) -> Iterator.Element? {
        return indices.contains(index) ? self[index] : nil
    }
}

Swift 3

extension Collection where Indices.Iterator.Element == Index {
    subscript (safe index: Index) -> Generator.Element? {
        return indices.contains(index) ? self[index] : nil
    }
}

Swift 2

extension Array {
    subscript (safe index: Int) -> Element? {
        return indices ~= index ? self[index] : nil
    }
}

• This way you’ll never hit Index out of range
• You’ll have to check if the item is nil

refer this question for more

---

Trying the Swift3 code in a Playground in Xcode 8.3.2 still leads to a
“crash” when I do let ar = [1,3,4], then let v = ar[5]. Why? – Thomas
Tempelmann May 17 at 17:40

You have to use our customized subscript so instead of let v = ar[5], it wll be let v = ar[safe: 5].

Default getting value from array.

let boo = foo[index]

Add use the customized subscript.

let boo = fee[safe: index]

// And we can warp the result using guard to keep the code going without throwing the exception.
guard let boo = foo[safe: index] else {
  return
}